Question

Find the number of ways in which 6 red roses and 3 white roses of different sizes can be made out to form a garland so that no two white roses come together.

Answer 1

Here, we need to find the number of ways in which the roses can be placed such that no two white roses are together. We will use the formula for circular permutations to find the number of ways in which the red roses can be placed. There will be 6 spaces between the 6 red roses, and we will need to find the number of ways to place the 3 white roses in those 6 spaces. Multiplying the number of ways in which the red roses can be placed and the number of ways to place the 3 white roses in the 6 spaces will give you the required answer.
Formula Used: We will be using two formulas here,

1. the formula for circular permutation, $\dfrac{{\left( {n - 1} \right)!}}{2}$, where $n$ is the number of objects.
2. the formula for permutations, ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$, where $r$ is the number of things to be chosen from $n$ number of sets.

Complete step by step solution:
Permutations are the number of ways in which objects of a set can be selected, where the order of objects is important.
The roses are of different sizes, and hence distinguishable from each other. Therefore, we will use permutation.
We know that the number of ways in which $n$ objects can be arranged in a circular form is $\dfrac{{\left( {n - 1} \right)!}}{2}$.
First, we need to arrange the red roses in a circular pattern.
Substituting 6 for $n$ in the formula for circular permutations, we get the number of ways to place the 6 red roses in a circle as
$\dfrac{{\left( {6 - 1} \right)!}}{2} = \dfrac{{5!}}{2}$
Simplifying this expression, we get
$\begin{array}{l}\dfrac{{\left( {6 - 1} \right)!}}{2} = \dfrac{{120}}{2}\\\ = 60\end{array}$
There are 60 ways in which the 6 red roses can be placed in a circle.
Now, no two white roses should be together. This means that there should be only one white rose between any two red roses.
We can see that there are 6 spaces between the 6 red roses when they are placed in a circle.
Hence, the 3 white roses can be placed in those 6 spaces.
The number of ways in which the 3 white roses can be placed in those 6 spaces can be found using the formula for permutations ${}^n{P_r} = \dfrac{{n!}}{{\left( {n - r} \right)!}}$.
Substituting 6 for $n$ and 3 for $r$, we get the number of ways to place the 3 white roses in the 6 spaces as
${}^6{P_3} = \dfrac{{6!}}{{\left( {6 - 3} \right)!}}$
Subtracting the terms in the denominator, we get
${}^6{P_3} = \dfrac{{6!}}{{3!}}$
Evaluating the factorial, we get
$\begin{array}{l}{}^6{P_3} = 6 \times 5 \times 4\\\ = 120\end{array}$
There are 120 ways in which the 3 white roses can be placed in those 6 spaces.
Now, we will multiply the number of ways in which the red roses can be placed in a circle and the number of ways to place the 3 white roses in the 6 spaces to get the number of ways to place all the roses in a circle, such that no two white roses are together.
Multiplying 60 and 120, we get
$60 \times 120 = 7200$

Therefore, the number of ways to place all the roses in a circle, such that no two white roses are together is 7200.

Note:
Permutation is a way of arranging elements of a set in a certain order or sequence. However, combination is a way of selecting elements from a set such that the order doesn’t matter. Since, in the question, the roses are of different sizes, and hence distinguishable from each other, we will use permutation. We will not use here the formula for combinations, that is ${}^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$. The combination formula gives the number of ways to place the 3 white roses in the 6 spaces as 20, which is incorrect.